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Intersection of Sets

Let the rando...

Question

Let the random variable X follow B(6, p). If 16 P(X = 4) = P(X = 2), then what is the value of p?

\frac{1}{3}

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\frac{1}{4}

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\frac{1}{5}

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\frac{1}{6}

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Solution

The correct option is C $\frac{1}{5}$
$16 P (X = 4) = P (X = 2)$

$∴ 16 (^{6} C_{4} p^{4} q^{2}) =^{6} C_{2} p^{2} q^{4}$

$∴ 16 (^{6} C_{4} p^{4} (1 - p)^{2}) =^{6} C_{2} p^{2} (1 - p)^{4}$

$∴ 16 p^{2} = (1 - p)^{2}$

$∴ 15 p^{2} + 2 p - 1 = 0$

$∴ (3 p + 1) (5 p - 1) = 0$

$∴ p = - \frac{1}{3}$ or $p = \frac{1}{5}$

Probability cannot be negative.
So, $p = \frac{1}{5}$

Answer is option (C)

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